LECTURE 8
Differentiation: Applications
1.) Revising the concept:
Differentiation helps to measure the gradient of a tangent to a curve at a particular point.
Example 1:
y = x 2 – 4x
Find the gradient of the tangent at:
i.) x = 1
ii.) x = 2
iii.) x = 3
Comment on your answers
Note that when dy/dx = 0, the tangent is parallel to the x-axis. Therefore the values of x when dy/dx = 0 show the turning points of the curve.
2.) Turning points:
The values of x where dy/dx = 0, indicate turning points. 2 types of turning points, maximum and minimum, can be identified by the following steps:
Step 1: Find dy/dx
Step 2: Find x when dy/dx = 0
Step 3: Find d2y/dx2
Step 4: Substitute x values found in step 2 into step 3
Note: a. When d2y/dx2 is positive (> 0), it represents a minimum point.
b. When d2y/dx2 is negative (<0), it represents a maximum point.
These maximum and minimum points are illustrated in the PowerPoint slides. There is a third type of turning point called the inflection point. This point is not relevant for the types of applications we consider below.
Example 2a:
Find the coordinates of all turning points and determine the nature of the turning points
y = x2 – 4x + 4
Example 2b:
Find the coordinates of all turning points and determine the nature of the turning points
y = 6x – x2
The maximum and minimum turning points are very useful in computing maximum revenues and profits and minimum costs. We examine some of these applications below
3.) Using differentiation to compute maximum revenues
Formula: REVENUES = PRICE (p) x QUANTITY (q)
Example 3a:
The demand function for a firm is given by q = 10,000 – 125p, where q is the number of units demanded and p is the price per unit.
Find:
(a.) The price that results in maximum revenue
(b.) The maximum revenue
(c.) The quantity sold to obtain this maximum revenue
4.) Finding minimum costs
Example 4:
A firm discovers that its total cost of purchasing, maintaining and storing inventory (total inventory cost) for a certain product is related to the order size, that is, the number of units it purchases in each order. This cost can be presented as follows:
C = (4860/q) + 15q + 750,000
where q is the order size, C is the total inventory cost.
Find:
(a.) The order size that minimizes total inventory cost
(b.) The expected minimum inventory cost
5.) Minimizing average cost per unit
FORMULA: AVERAGE COST (AC) = TOTAL COST (TC)/QUANTITY (q)
Example 5:
The total cost of producing q units of a certain product is given by:
C = 100,000 + 1500q + 0.2 q2
where C = total cost.
a.) Find the value of q where the average cost per unit is minimized.
b.) How much is the minimum average cost?
6.) Maximizing profit, given demand and cost functions
FORMULA:
PROFIT (p) = TOTAL REVENUE (TR) – TOTAL COST (TC)
Example 6:
A manufacturer has compiled the following information regarding a new product:
Estimated demand function:
q = 100,000 – 200p, where q = estimated number of units demanded at a price p
Estimated cost function:
C = 150,000 + 100q + 0.003q2
(a.) Formulate
an equation for the profit, p
(b.) At what values of q will profit be maximized?
(c.) Find the estimated maximum profit
7.) The marginal cost/revenue approach
FORMULA: (TO MAXIMISE PROFIT)
MARGINAL REVENUE (MR) = MARGINAL COST (MC)
Given a cost function, C, stated in terms of q, MC = dC/dq
Given a revenue function, R, stated in terms of q, MR = dR/dq
Example 7:
Find the maximum profit in Example 6 using the marginal approach
Note:
Revenue, R = 500q – 0.005q2
Cost, C = 150,000 + 100q + 0.003q2
Independent reading: BUD, Chapters 16, 17
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